Which of the following is the maximum value of the function f(x)=2sinxcosx?

a) 0

b) 1

c) 2

d) No maximum value

since 2sinx cosx = sin(2x), it has a max of 1. sin(u) has a min/max of 1 regardless of what u is.

To find the maximum value of the function f(x) = 2sin(x)cos(x), we can use trigonometric identities. Specifically, we'll use the double angle identity: sin(2x) = 2sin(x)cos(x).

Let's rewrite the function using the double angle identity:
f(x) = sin(2x)

The maximum value of sin(2x) is 1, which occurs when 2x = π/2.

To determine the maximum value of f(x), we need to find the corresponding value of x.
Solving for x, we have:
2x = π/2
x = π/4

So, the maximum value of f(x) is f(π/4) = sin(2(π/4)) = sin(π/2) = 1.

Therefore, the correct answer is b) 1.

To find the maximum value of the function f(x) = 2sin(x)cos(x), we need to look for the critical points by finding the first derivative of the function.

First, let's find the derivative of f(x):

f'(x) = d/dx (2sin(x)cos(x))

Using the product rule, we can differentiate the function:

f'(x) = 2(cos^2(x) - sin^2(x))

Now, to find the critical points, we set f'(x) = 0:

0 = 2(cos^2(x) - sin^2(x))

Since the derivative equals zero when the expression inside the parentheses equals zero:

0 = cos^2(x) - sin^2(x)

Using the trigonometric identity for the difference of squares:

0 = cos(2x)

So, the critical points occur when cos(2x) = 0.

The solutions to cos(2x) = 0 are x = π/4 + nπ/2, where n is an integer.

Now, let's investigate the behavior of the function at the critical points.

To do this, we analyze the second derivative of f(x):

f''(x) = d²/dx² (2sin(x)cos(x))

Using the product rule again:

f''(x) = 2(-sin^2(x) - cos^2(x))

Simplifying:

f''(x) = -2

Since f''(x) is a constant -2, we know it is negative for all x.

This means the function f(x) = 2sin(x)cos(x) is concave down and has a maximum value at x = π/4 + nπ/2.

Now, we need to substitute these values into the original function f(x) to find the maximum value.

f(π/4 + nπ/2) = 2sin(π/4 + nπ/2)cos(π/4 + nπ/2)

Since sin(π/4) = cos(π/4) = 1/√2, this simplifies to:

f(π/4 + nπ/2) = 2(1/√2)(1/√2) = 2/2 = 1

Therefore, the maximum value of the function f(x) = 2sin(x)cos(x) is 1.

Therefore, the correct answer is (b) 1.